Problem: 2 diagonals of a regular heptagon (a 7-sided polygon) are chosen.  What is the probability that they intersect inside the heptagon?
Solution: There are $\binom{7}{2} = 21$ pairs of points in the heptagon, and all but 7 (the sides of the heptagon) are diagonals, which means there are 14 diagonals.  So there are $\binom{14}{2} = 91$ pairs of diagonals.  Any four points on the heptagon uniquely determine a pair of intersecting diagonals. (If vertices $A,B,C,D$ are chosen, where $ABCD$ is a convex quadrilateral, the intersecting pair of diagonals are $AC$ and $BD$.)  So the number of sets of intersecting diagonals is the number of combinations of 4 points, which is $\binom{7}{4} = 35$.  So the probability that a randomly chosen pair of diagonals intersect is $\dfrac{35}{91} = \boxed{\dfrac{5}{13}}$.